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M13/5/MATSD/SP2/ENG/TZ2/XX/M

24 pages

MARKSCHEME

May 2013

MATHEMATICAL STUDIES

Standard Level

Paper 2

– 2 – M13/5/MATSD/SP2/ENG/TZ2/XX/M

This markscheme is confidential and for the exclusive use of

examiners in this examination session.

It is the property of the International Baccalaureate and must not

be reproduced or distributed to any other person without the

authorization of the IB Assessment Centre.


Paper 2 Markscheme

Instructions to Examiners

Notes: If in doubt about these instructions or any other marking issues, contact your team leader

for clarification.

1 Abbreviations

M Marks awarded for Method

A Marks awarded for an Answer or for Accuracy

R Marks awarded for clear Reasoning

G Marks awarded for correct solutions obtained from a Graphic Display Calculator, when no

working shown.

AG Answer Given in the question and consequently, marks not awarded.

ft Marks that can be awarded as follow through from previous results in the question.

2 Method of Marking

(a) All marking must be done in scoris using the mathematical studies annotations and in

accordance with the current document for guidance in e-marking Mathematical Studies SL.

It is essential that you read this document before you start marking.

(b) If a question part is completely correct use the number tick annotations to award full marks.

If a part is completely wrong use the A0 annotation, otherwise full annotations must be shown.

(c) Working crossed out by the candidate should not be awarded any marks.

(d) Where candidates have written two solutions to a question, only the first solution should be marked.

(e) If correct working results in a correct answer but then further working is developed, full marks may

not always be awarded. Full marks will be awarded if the candidate shows correct working leading to

the correct answer. See also section 4(c).

Example: Calculate the gradient of the line passing through the points (5, 3) and (0, 9) .

Markscheme Candidates’ Scripts Marking

9 3

0 5

(M1)

Award (M1) for correct

substitution in gradient

formula

6

5 (A1)

(i) 9 3 6

0 5 5

(M1)

Gradient is 6

5 (A1)

(There is clear understanding of the gradient.)

69

5y x

(ii) 9 3 6

0 5 5

(M1)

6

95

y x (A0)

(There is confusion about what is required.)


3 Follow-through (ft) Marks

Errors made at any step of a solution affect all working that follows. To limit the severity of the penalty,

follow through (ft) marks can be awarded. Markschemes will indicate where it is appropriate to apply

follow through in a question with ‘(ft)’.

(a) Follow through applies only from one part of a question to a subsequent part of the question.

Follow through does not apply within the same part.

(b) If an answer resulting from follow through is extremely unrealistic (e.g. negative distances or

incorrect by large order of magnitude) then the final A mark should not be awarded.

(c) If a question is transformed by an error into a different, much simpler question then follow through

may not apply.

(d) To award follow through marks for a question part, there must be working present for that part.

An isolated follow through answer, without working is regarded as incorrect and receives no marks

even if it is approximately correct.

(e) The exception to the above would be in a question which is testing the candidate’s use of the GDC,

where working will not be expected. The markscheme will clearly indicate where this applies.

(f) Inadvertent use of radians will be penalised the first time it occurs. The markscheme will give clear

instructions to ensure that only one mark per paper can be lost for the use of radians.

Example: Finding angles and lengths using trigonometry


(a) sin sin30

3 4

A (M1)(A1)

Award (M1) for substitution in sine

rule formula, (A1) for correct

substitutions.

22.0 (22.0243 )A (A1)(G2)

(b) 7tan (22.0243 )x (M1)

2.83 (2.83163 ) (A1)(ft)

(a) sin sin30

4 3

A

(M1)(A0)

(use of sine rule but

with wrong values)

41.8A (A0)

(Note: the 2nd (A1) here was not marked (ft)

and cannot be awarded because there was

an earlier error in the same question part.)

(b) case (i) 7tan 41.8x (M1)

6.26 (A1)(ft)

but case (ii) 6.26 (G0)

since no working shown


4 Using the Markscheme

(a) A marks are dependent on the preceding M mark being awarded, it is not possible to

award (M0)(A1). Once an (M0) has been awarded, all subsequent A marks are lost in that part of

the question, even if calculations are performed correctly, until the next M mark.

The only exception to this will be for an answer where the accuracy is specified in the question – see

section 5.

(b) A marks are dependent on the R mark being awarded, it is not possible to award (A1)(R0). Hence the

(A1) cannot be awarded for an answer which is correct when no reason or the wrong reason is given.

(c) In paper 2 candidates are expected to demonstrate their ability to communicate mathematics using

appropriate working. Answers which are correct but not supported by adequate working will

not always receive full marks, these unsupported answers are designated G in the mark scheme as an

alternative to the full marks. Example (M1)(A1)(A1)(G2).

Example: Using trigonometry to calculate an angle in a triangle.


(a) sin sin30

3 4

A (M1)(A1)

Award (M1) for substitution in sine

rule formula, (A1) for correct

substitutions.

22.0 (22.0243 )A (A1)(G2)

(i) sin sin30

3 4

A

(M1)(A1)

22.0A (A1)

(ii) 22.0A (G2)

Note: G marks are used only if no working has been shown

and the answer is correct.

(d) Alternative methods may not always be included. Thus, if an answer is wrong then the working must

be carefully analysed in order that marks are awarded for a different method consistent with the

markscheme.

Where alternative methods for complete questions are included in the markscheme, they are indicated

by ‘OR’ etc.

(e) Unless the question specifies otherwise, accept equivalent forms. For example: sin

cos

for tan .

On the markscheme, these equivalent numerical or algebraic forms will sometimes be written in

brackets after the required answer.

Where numerical answers are required as the final answer to a part of a question in the markscheme,

the scheme will show, in order:

the 3 significant figure answer worked through from full calculator display;

the exact value (for example 3 if applicable);

the full calculator display in the form 2.83163… as in the example above.

Where answers are given to 3 significant figures and are then used in subsequent parts of the question

leading to a different 3 significant figure answer, these solutions will also be given.


(f) As this is an international examination, all valid alternative forms of notation should be accepted.

Some examples of these are:

Decimal points: 1.7; 1’7; 1 7 ; 1,7 .

Different descriptions of an interval: 3 < x < 5; (3, 5); ] 3, 5 [ .

Different forms of notation for set properties (e.g. complement): ; ; ; ;(cA A A U A A ;U \ A.

Different forms of logic notation: p ; p ; p ; p ; ~ p.

p q ; p q ; q p .

(g) Discretionary marks: There will be very rare occasions where the markscheme does not cover the

work seen. In such cases the annotation DM should be used to indicate where an examiner has used

discretion. Discretion should be used sparingly and if there is doubt and exception should be raised

through scoris to the team leader.


There will be no whole paper penalty marks for accuracy AP, financial accuracy FP and units UP. Instead

these skills will be assessed in particular questions and the marks applied according to the rules given in

sections 5, 6 and 7 below.

5 Accuracy of Answers

Incorrect accuracy should be penalized once only in each question according to the rules below.

Unless otherwise stated in the question, all numerical answers should be given exactly or correct to

3 significant figures.

1. If the candidate’s unrounded answer is seen and would round to the required 3 sf answer, then award

(A1) and ignore subsequent rounding.

2. If the candidate’s unrounded answer is not seen then award (A1) if the answer given is correctly

rounded to 2 or more significant figures, otherwise (A0).

Note: If the candidate’s unrounded answer is not seen and the answer is given correct to 1 sf (correct or not),

the answer will be considered wrong and will not count as incorrect accuracy. If this answer is used in

subsequent parts, then working must be shown for further marks to be awarded.

3. If a correct 2 sf answer is used in subsequent parts, then working must be shown for further marks to

be awarded. (This treatment is the same as for following through from an incorrect answer.)

These 3 points (see numbers in superscript) have been summarized in the table below and illustrated in the

examples following.

If candidates final answer is given …

Exact or

correct to

3 or more sf

Incorrect to

3 sf

Correct to

2 sf3 Incorrect to

2 sf

Correct or

incorrect to 1 sf

Unrounded

answer seen1 Award the final (A1) irrespective of correct or incorrect rounding

Unrounded

answer not seen2 (A1) (A0) (A1) (A0) (A0)

Treatment of

subsequent parts As per MS Treat as follow through, only if working is seen3


Examples:


9.43 (9.43398 ) (A1)

(i) 9.43398 is seen followed by 9; 9.4;

9.43; 9.434 etc. (correctly rounded) (A1)

(ii) 9.43398 is seen followed by 9.433;

9.44 etc. (incorrectly rounded) (A1)

(iii) 9.4 (A1)

(iv) 9 (A0)

(correct to 1 sf)

(v) 9.3 (A0)

(incorrectly rounded to 2 sf)

(vi) 9.44 (A0)



7.44 (7.43798 ) (A1)

(i) 7.43798 is seen followed by 7; 7.4;

7.44; 7.438 etc. (correctly rounded) (A1)

(ii) 7.43798 is seen followed by 7.437;

7.43 etc. (incorrectly rounded) (A1)

(iii) 7.4 (A1)

(iv) 7 (A0)

(correct to 1 sf)

(v) 7.5 (A0)


(vi) 7.43 (A0)



Example: ABC is a right angled triangle with angle ABC 90 , AC 32 cm and AB 30 cm . Find (a) the

length of BC, (b) The area of triangle ABC.


(a) 2 2BC 32 30 (M1)

Award (M1) for correct substitution in

Pythagoras’ formula

11.1 124,11.1355... (cm) (A1)

(b) 1

Area 30 11.1355... 2

(M1)

Award (M1) for correct substitution in

area of triangle formula

2167(167.032...)(cm ) (A1)(ft)

(a) 2 2BC 32 30 (M1)

11(cm) (A1)

(2 sf answer only seen, but correct)

(b) case (i) 1

Area 30 112

(M1)

(working shown)

2165 (cm ) (A1)(ft)

case (ii) 2165 (cm ) (M0)(A0)(ft)

(No working shown, the answer 11 is

treated as a ft, so no marks awarded here)

Certain answers obtained from the GDC are worth 2 marks and working will not be seen. In these cases only

one mark should be lost for accuracy.

e.g. Chi-squared, correlation coefficient, mean


Chi-squared

7.68 (7.67543 ) (A2)

(a) 7.7 (G2)

(b) 7.67 (G1)

(c) 7.6 (G1)

(d) 8 (G0)

(e) 7 (G0)

(e) 7.66 (G0)


Regression line


0.888 13.5 y x (A2) ( 0.887686 13.4895 ) y x

If an answer is not in the form of

an equation award at most

(A1)(A0).

(a) 0.89 13y x (G2)

(both accepted)

(b) 0.88 13 y x (G1)

(one rounding error)

(c) 0.88 14 y x (G1)

(rounding error repeated)

(d) (i) 0.9 13 y x

(ii) 0.8 13 y x (G1)

(1 sf not accepted)

(e) 0.88 13x (G0)

(one rounding error and not an equation)

Maximum/minimum/points of intersection


(2.06, 4.49) (A1)(A1)

(2.06020 , 4.49253 )

(a) (2.1, 4.5) (A1)(A1)

(both accepted)

(b) (2.0, 4.4) (A1)

(same rounding error twice)

(c) (2.06, 4.4) (A1)

(one rounding error)

(d) (2, 4.4) (A0)

(1sf not accepted, one rounding error)

Rounding of an exact answer to 3 significant figures should be accepted if performed correctly.

Exact answers such as 1

4 can be written as decimals to fewer than three significant figures if the result is

still exact. Reduction of a fraction to its lowest terms is not essential, however where an answer simplifies to

an integer this is expected.

Ratios of π and answers taking the form of square roots of integers or any rational power of an integer

(e.g. 23 413, 2 , 5 ,) may be accepted as exact answers. All other powers (e.g. of non-integers) and values

of transcendental functions such as sine and cosine must be evaluated.

If the level of accuracy is specified in the question, a mark will be allocated for giving the answer to

the required accuracy. In all such cases the final mark is not awarded if the rounding does not follow the

instructions given in the question. A mark for specified accuracy can be regarded as a (ft) mark regardless of

an immediately preceding (M0).


6 Level of accuracy in finance questions

The accuracy level required for answers will be specified in all questions involving money. This will usually

be either whole units or two decimal places. The first answer not given to the specified level of accuracy

will not be awarded the final A mark. The markscheme will give clear instructions to ensure that only

one mark per paper can be lost for incorrect accuracy in a financial question.

Example: A financial question demands accuracy correct to 2 dp.


$231.62 (231.6189) (A1)

(i) 231.6 (A0)

(ii) 232 (A0)

(Correct rounding to incorrect level)

(iii) 231.61 (A0)

(iv) 232.00 (A0)

(Parts (iii) and (iv) are both

incorrect rounding to correct level)

7 Units in answers

There will be specific questions for which the units are required and this will be indicated clearly

in the markscheme. The first correct answer with no units or incorrect units will not be awarded the

final A mark. The markscheme will give clear instructions to ensure that only one mark per paper can be

lost for lack of units or incorrect units.

The units are considered only when the numerical answer is awarded (A1) under the accuracy rules given

in Section 5.

Example:


(a) 237000 m (A1)

(b) 33200 m (A1)

(a) 236000 m (A0)

(Incorrect answer so units not considered)

(b) 23200 m (A0)

(Incorrect units)

If no method is shown and the answer is correct but with incorrect or missing units award G marks

with a one mark penalty.

8 Graphic Display Calculators

Candidates will often be obtaining solutions directly from their calculators. They must use mathematical

notation, not calculator notation. No method marks can be awarded for incorrect answers supported only by

calculator notation. The comment ‘I used my GDC’ cannot receive a method mark.


QUESTION 1

(a)

463

1

72 5

U

C M

P

(A1)(A1)(A1)(A1) [4 marks]

Award (A1) for 3 intersecting circles and rectangle, (A1) for 1, 3, 4 and 7,

(A1) for 2, (A1) for 6 and 5.

(b) (i) 2 (A1)(ft)

(ii) 6 (A1)(ft)

(iii) 40 1 6 2 3 4 7 5 (M1)

12 (A1)(ft)(G2) [4 marks]

(c) (i) 16 2

, 0.4, 40 %40 5

(A1)(A1)(G2)

(ii) 20 1

, 0.5, 50 %40 2

(A1)(ft) (A1) (G2)

continued…

Note: Award (A1)(ft) for numerator, (A1) for denominator. Follow through from their Venn

diagram. Answer must be less than 1 otherwise award (A0)(A0). Award (A0)(A0) if

answer is given as incorrect reduced fraction without working.

Note: Award (M1) for subtracting all their values from 40.

Note: Follow through from their Venn diagram for parts (i), (ii) and (iii).

Note: Award (A1) for numerator, (A1) for denominator. Answer must be less than 1 otherwise

award (A0)(A0). Award (A0)(A0) if answer is given as incorrect reduced fraction without

working.


Question 1 continued

(iii) 6 3

, 0.15,15 %40 20

(A1)(ft)(A1)(G2)

(iv) 11

(0.6875, 68.75 %)16

(A1)(ft)(A1)(G2) [8 marks]

(d) 16 15

40 39 (A1)(A1)(ft)

240 2

,0.153846 ,15.4 %1560 13

(A1)(ft)(G2) [3 marks]

Total [19 marks]




Note: Follow through from their answer to part (c)(i).

Answer must be less than 1 otherwise award at most (A1)(A1)(A0)(ft).

Note: Award (A1) for multiplication of their probabilities, (A1)(ft) for their correct probabilities.





QUESTION 2

(a)

(A1) for correct scales and labels (mass or m on the horizontals axis, time or t

on the vertical axis)

(A3) for 7 or 8 correctly placed data points

(A2) for 5 or 6 correctly placed data points

(A1) for 3 or 4 correctly placed data points, (A0) otherwise. (A4) [4 marks]

(b) (i) 1.91 (kg) (1.9125 kg) (G1)

(ii) 83 (minutes) (G1) [2 marks]

(c) Their mean point labelled. (A1)(ft) [1 mark]

(d) Line of best fit drawn on scatter diagram. (A1)(ft)(A1)(ft) [2 marks]

continued…

Note: If axes reversed award at most (A0)(A3)(ft). If graph paper not used, award

at most (A1)(A0).

Notes:Award (A1)(ft) for straight line through their mean point, (A1)(ft) for line of

best fit with intercept 9( 2) . The second (A1)(ft) can be awarded even if the

line does not reach the t-axis but, if extended, the t-intercept is correct.

Note: Follow through from part (b). Accept any clear indication of the mean point.

For example: circle around point, (m, t), M , etc.



(e) 75 (M1)(A1)(ft)(G2) [2 marks]

(f) 0.960 (0.959614 ) (G2) [2 marks]

(g) Strong and positive (A1)(ft)(A1)(ft) [2 marks]

(h) (i) Cooking time is much larger (or smaller) than the other eight (A1)

(ii) The gradient of the new line of best fit will be larger (or smaller) (A1) [2 marks]

Total [17 marks]

Note: Some acceptable explanations may include but are not limited to:

The line of best fit may be further away from the plotted points

It may be steeper than the previous line (as the mean would change)

The t-intercept of the new line is smaller (larger)

Do not accept vague explanations, like:

The new line would vary

It would not go through all points

It would not fit the patterns

The line may be slightly tilted

Note: Award (G0)(G1)(ft) for 0. 95, 0.959

Note: Follow through from their correlation coefficient in part (f).

Notes: Accept 74.77 from the regression line equation.

Award (M1) for indication of the use of their graph to get an estimate

OR for correct substitution of 1.7 in the correct regression line

equation 38.5 9.32t m .


QUESTION 3 Units are required in parts (b), (c), (d) and (e). The first time the correct units are

not given the final (A1) is not awarded; do not penalize more than once.

(a) 1.6

STcos35

(M1)(A1)

OR

ST 3.2

sin35 sin110 (M1)(A1)

ST 1.95323 (A1)

1.95 (m) (AG) [3 marks]

(b) 1

3.2 1.95323 sin352

or 1

1.95323 1.95323 sin1102 (M1)(A1)

21.79 m 2(1.79253 m ) (A1)(G2)

OR

13.2 1.12033...

2 (A1) (M1)

21.79 m 2(1.79253 m ) (A1)(G2) [3 marks]

continued…

Notes: The answer is 21.79 m , units are required. Accept 1.78955 from using 1.95.

Note: Award (M1) for substituted area formula, (A1) for correct substitutions. Do not award

follow through marks.

Note: Award (A1) for the correct value for TM (1.12033...) OR correct expression for TM

(i.e. 1.6tan35 , 2 21.95323... 1.6 ), (M1) for correctly substituted formula for triangle

area.

Notes: The answer is 21.79 m , units are required. Accept 1.78 m 2 from using 1.95.

Notes: Both unrounded and rounded answer must be seen for final (A1) to be awarded.

Note: Award (M1) for substituted sine rule equation, (A1) for correct substitutions.

Note: Award (M1) for correctly substituted trig equation, (A1) for 1.6 seen.



(c) 2 29.18 m (9.18022 m ) (A1)(G1) [1 mark]

(d) 2 1.79253 2 9.18022 4.7 3.2 (M1)(A1)(ft)

2 237.0 m (36.9855 m ) (A1)(ft)(G2) [3 marks]

(e) 1.79253 4.7 (M1)

3 38.42 m (8.42489 m ) (A1)(ft)(G2) [2 marks]

continued…

Notes: The answer is 38.42 m , units are required. Accept 38.41m from use of 1.79.

An answer of 8.35, from use of TM 1.11 , will receive follow-through marks if

working is shown. Follow through from their answer to part (b). Do not penalize if

lack of units was penalized earlier in the question.

Note: Award (M1) for their correctly substituted volume formula.

Notes: The answer is 237.0 m , units are required. Accept 236.98 m from using 3sf

answers.

Follow through from their answers to (b) and (c).

Do not penalize if lack of units was penalized earlier in the question.

Note: Award (M1) for addition of three products, (A1)(ft) for three correct products.

Notes: The answer is 29.18 m , units are required. Do not penalize if lack of units was

already penalized in (b). Do not award follow through marks here.

Accept 2 29.17 m (9.165 m ) from using 1.95.



(f) (i) TM 1.6 tan35 (M1)

OR

2 2TM 1.95323... 1.6 (M1)

OR

3.2 TM

1.79253...2

(M1)

1.12 (m) (1.12033 ) (A1)(ft)(G2)

(ii) 2 2VM 1.12033 4.7 (M1)

4.83 (m) (4.83168 ) (A1)(ft)(G2) [4 marks]

continued…

Note: Award (M1) for their correct substitution in area of triangle formula.

Note: Award (M1) for correct substitution in Pythagoras’ theorem.

Notes: Follow through from (f)(i).

Note: Award (M1) for their correct substitution in Pythagoras’ theorem.

Notes: Follow through from their answer to (b) if area of triangle is used.

Accept 1.11 (1.11467) from use of ST 1.95 .

Notes: Award (M1) for their correct substitution in trig ratio.



(g) 1 1.12033sin

4.83168

(M1)

OR

1 4.7cos

4.83168

(M1)

OR

1 1.12033tan

4.7

(M1)

OR

2 22

14.7 4.83168... 1.12033...

cos2 4.7 4.83168...

(M1)

13.4 (13.4073 ) (A1)(ft)(G2) [2 marks]

Total [18 marks]

Note: Award (M1) for correctly substituted cosine formula.

Notes: Accept 13.3 . Follow through from part (f).

Note: Award (M1) for correctly substituted trig equation.


QUESTION 4

(a) 120 10 4 (M1)(A1)

160 (A1)(G3) [3 marks]

(b) 120 ( 1) 10 260n (M1)(M1)

15 (A1)(G2) [3 marks]

(c) 15

(120 260)2

or 15

(2 120 (15 1) 10)2

(M1)(A1)(ft)

2850 seconds (A1)(ft)(G2)

47.5 minutes (A1)(ft)(G3) [4 marks]

(d) 3 1120 1.06 (M1)(A1)

135 (134.832) (A1)(G2) [3 marks]

(e) 4

4

120(1.06 1)

(1.06 1)S

(M1)(A1)

525 (524.953 ) (A1)(G2) [3 marks]

continued…

Note: Award (G2) for 2850 seen with no working shown.

Notes: Award (M1) for substituted GP sum formula, (A1) for correct

substitutions.

Accept a sum of a list of 4 correct terms.

Notes: Award (M1) for substituted GP formula, (A1) for correct substitutions.

Accept a list of 3 correct terms.

Notes: A final (A1)(ft) can be awarded for correct conversion from seconds

into minutes of their incorrect answer.

Follow through from their answer to part (b).

Notes: Award (M1) for substituted AP sum formula, (A1)(ft) for correct

substitutions.

Accept a sum of a list of 15 correct terms.

Follow through from their answer to part (b).

Notes: Award (M1) for correctly substituted AP formula, (M1) for equating

to 260.

Accept a list of correct terms showing at least the 14th and 15th terms.

Notes: Award (M1) for substituted AP formula, (A1) for correct substitutions.

Accept a list of 4 correct terms.



(f) 1120 ( 1) 10 120 1.06nn (M1)(M1)

OR

List of at least 2 terms for both sequences (120, 130, … and 120, 127.2, …) (M1)

List of correct 12th and 13th terms for both sequences (..., 230, 240 and …,

227.8, 241.5) (M1)

OR

A sketch with a line and an exponential curve, (M1)

An indication of the correct intersection point (M1)

13th lap (A1)(ft)(G2) [3 marks]

Total [19 marks]

Note: Do not award the final (A1)(ft) if final answer is not a positive integer.

Notes: Award (M1) for correct left hand side, (M1) for correct right hand side.

Accept an equation.

Follow through from their expressions given in parts (a) and (d).


QUESTION 5

(a) 275 27

7510 2

y (M1)

450 (A1)

Yes, point A is on the bike track. (A1) [3 marks]

(b) d 2 27

d 10 2

y x

x

d0.2 13.5

d

yx

x

(A1)(A1) [2 marks]

(c) 2 27

010 2

x (M1)

67.5x (A1)(ft)

(Their) 67.5 75 (R1)

continued…

Note: Award (R1) for a comparison of their 67.5 with 75. Comparison may be implied

(eg 67.5 is the x-coordinate of the furthest north point).

Note: Follow through from their derivative from part (b).

Note: Award (M1) for equating their derivative from part (b) to zero.

Notes: Award (A1) for each correct term. If extra terms are seen award at

most (A1)(A0). Accept equivalent forms.

Note: Do not award the final (A1) if correct working is not seen.

Note: Award (M1) for substitution of 75 in the formula of the function.



OR

d 2 (75) 27

d 10 2

y

x

(M1)

1.5 (A1)(ft)

(Their) 1.5 0 (R1)

Hence A is not the furthest north point. (A1)(ft) [4 marks]

(d) (i) M(50,175) (A1)

(ii) 350 0

100 0

(M1)

350 7

3.5 ,100 2

(A1)(ft)(G2) [3 marks]

continued…

Note: Follow through from (d)(i) if midpoint is used to calculate gradient. Award

(G1)(G0) for answer 3.5x without working.

Note: If parentheses are omitted award (A0).

Accept 50, 175x y .

Note: Award (R1) for a comparison of their –1.5 with 0. Comparison may be implied

(eg The gradient of the parabola at the furthest north point (vertex) is 0).

Note: Award (M1) for correct substitution in gradient formula.

Note: Do not award (R0)(A1)(ft). Follow through from their derivative from part (b).

Note: Follow through from their derivative from part (b).

Note: Award (M1) for substitution of 75 into their derivative from part (b).



(e) 3.5 150y x (A1)(ft)(A1)(ft)

3.5 150x y or 7 2 300x y (or equivalent) (A1)(ft) [3 marks]

(f) (18.4, 214) (18.3772 , 214.320 ) (A1)(ft)(A1)(ft)(G2)(ft) [2 marks]

Total [17 marks]

Notes: Follow through from their equation in (e). Coordinates must be positive for follow

through marks to be awarded.

If parentheses are omitted and not already penalized in (d)(i) award at most

(A0)(A1)(ft).

If coordinates of the two intersection points are given award (A0)(A1)(ft).

Accept 18.4x , 214y .

Note: Award (A1)(ft) for expressing their equation in the form

a b cx y .

Note: Award (A1)(ft) for using their gradient from part (d), (A1)(ft) for

correct equation of line.

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